# Taylor error bound for non-alternating series [closed]

So Brilliant.org says this: Since the difference between Pn and Pn+1 is just that last term [of Pn+1], the error of Pn can be no larger than that term. I get how this works for alternating series, because the polynomial "crosses" over the function value every time you add a term (if function value is 5, it might go 1, then 7, 4, then 5.5), so your the difference from that term is greater than the difference of your polynomial from the root function. But what about positive series like e^x?  With that, if your actual value is 5, you get 1, then 3, then 4, then 4.5, etc.  So the difference between a four term and a five term polynomial is less than the difference between the four term and the actual function.  So clearly the error is greater than that.   Am I missing some way that this is somehow correct?  Or are they wrong in that particular case?  and if so, is there a way to actually find the error bound?

I have to teach a lesson on taylor series error bound in a few days and not understanding this part is really bugging me.

Thanks!

## closed as unclear what you're asking by JonMark Perry, user284331, cansomeonehelpmeout, Chris Custer, José Carlos SantosApr 12 '18 at 21:36

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This is just nonsense. Consider $$\sin(t)=1-\frac{t^2}{2!}+\dots$$
In this example $P_4=P_5$, so Brilliant.org's claim would imply that $P_4(t)=\sin(t)$.
No, in general the bound is not given by the next term. The Lagrange error formula however says that the error is "like" the next term, in that it is the next term but with the derivative evaluated at a point between the center of expansion $a$ and the evaluation point $x$. But the difference could be considerable if that derivative changes a lot between $x$ and $a$.
For example, for $e^x,x>0$ one can show that $e^x-T_n(x)<e^x x^{n+1}/(n+1)!$.